Question

Phosphate is precipitated from its solution with ammonium molybdate, as (NH4)3[PMo12O40XH₂0].

Since the precipitate does not have a constant composition with regard to water content, it is dissolved
in ammonia and the molybdate is precipitated with Pb(NO3)2, as PbMoO4.
If the final precipitate from PbMoO4, weighs 1.100 g, what is the weight of P in the initial sample?

Answer 1

To determine the weight of P in the initial sample, we can set up a proportion based on the reaction equations.

The molar ratio between (NH4)3[PMo12O40XH₂0] and PbMoO4 is 3 moles of P per 1 mole of PbMoO4. We can calculate the molar mass of PbMoO4, which is 303.238 g/mol.

1.100 g of PbMoO4 is equivalent to 0.00363 moles (1.100 g / 303.238 g/mol).

Since there are 3 moles of P in (NH4)3[PMo12O40XH₂0], the initial sample must contain 3 times more moles of P.

Therefore, the weight of P in the initial sample is 3 times 0.00363 moles, which is approximately 0.01089 moles.

To calculate the weight of P, we multiply the moles by its molar mass. The molar mass of P is 30.97 g/mol.

Weight of P in the initial sample = 0.01089 moles * 30.97 g/mol ≈ 0.337 g

So, the weight of P in the initial sample is approximately 0.337 grams